Polynomial proof question.

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Problem.

Let $f(x)$ be a polynomial with real coefficients such that $f(x)>0$ for all real $x$. Prove that there are real polynomials $p(x)$ and $q(x)$ such that $f(x)=p^2(x)+q^2(x)$

My solution attempt:

The polynomial has no real roots since $f(x)>0$ for all real x. Suppose $f(x)$ has corresponding complex roots and count them as pairs

$$z_1,\bar{z_1},...,z_l,\bar{z_1}$$ And clearly $deg f=2l$ in this case. Now suppose $z_j=a_j+ib_j$ for $j=1,...,l.$ Then by the fundamental theorem of algebra
$f(x)$
$$f(x)=a_n(x-z_1)(x-\bar{z_1})\cdot\cdot\cdot(x-z_l)(x-\bar{z_l})$$$$f(x)=a_n(x^2-(z_1+\bar{z_1})x+z_1\bar{z_1})\cdot\cdot\cdot(x^2-(z_l+\bar{z_l})x+z_l\bar{z_l})$$$$f(x)=a_n(x^2-2a_1x+a_1^2+b_1^2)\cdot\cdot\cdot(x^2-2a_lx+a_l^2+b_l^2)$$
Hence $f(x)$ can be prime factored into real second-degree polynomials of the form

$$f(x)=a_n\prod_{j=1}^{l}(x-a_j)^2+b_j^2$$
And we notice that each term is always positive because it's a sum of squares, therefore $f(x)>0$ provided $a_n>0$. However, there remains to prove that the expansion of this expression can be written as $p^2(x)+q^2(x)$ and I'm not sure how to continue.

 

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Problem.

Let $f(x)$ be a polynomial with real coefficients such that $f(x)>0$ for all real $x$. Prove that there are real polynomials $p(x)$ and $q(x)$ such that $f(x)=p^2(x)+q^2(x)$

My solution attempt:

The polynomial has no real roots since $f(x)>0$ for all real x. Suppose $f(x)$ has corresponding complex roots and count them as pairs

$$z_1,\bar{z_1},...,z_l,\bar{z_1}$$ And clearly $deg f=2l$ in this case. Now suppose $z_j=a_j+ib_j$ for $j=1,...,l.$ Then by the fundamental theorem of algebra
$f(x)$
$$f(x)=a_n(x-z_1)(x-\bar{z_1})\cdot\cdot\cdot(x-z_l)(x-\bar{z_l})$$$$f(x)=a_n(x^2-(z_1+\bar{z_1})x+z_1\bar{z_1})\cdot\cdot\cdot(x^2-(z_l+\bar{z_l})x+z_l\bar{z_l})$$$$f(x)=a_n(x^2-2a_1x+a_1^2+b_1^2)\cdot\cdot\cdot(x^2-2a_lx+a_l^2+b_l^2)$$
Hence $f(x)$ can be prime factored into real second-degree polynomials of the form

$$f(x)=a_n\prod_{j=1}^{l}(x-a_j)^2+b_j^2$$
And we notice that each term is always positive because it's a sum of squares, therefore $f(x)>0$ provided $a_n>0$. However, there remains to prove that the expansion of this expression can be written as $p^2(x)+q^2(x)$ and I'm not sure how to continue.

One observation is that each term in the product has the form
$$(x-a_j)^2+b_j^2=(x-a_j+ib_j)(x-a_j-ib_j)=\lvert x-(a_j+ib_j) \rvert^2$$which is the magnitude squared of the complex number $x-a_j-ib_j$. Hence each factor $(x-a_j)^2+b_j^2$ can be rewritten using the modulus of a complex number.

 

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One observation is that each term in the product has the form
$$(x-a_j)^2+b_j^2=(x-a_j+ib_j)(x-a_j-ib_j)=\lvert x-(a_j+ib_j) \rvert^2$$which is the magnitude squared of the complex number $x-a_j-ib_j$

The constant $a_n$ is included to ensure scaling of $f(x)$ and each factor $x-(a_j+ib_j)$ is a linear term in $x$, involving the complex roots $a_j+ib_j$.

 

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The constant $a_n$ is included to ensure scaling of $f(x)$ and each factor $x-(a_j+ib_j)$ is a linear term in $x$, involving the complex roots $a_j+ib_j$.

Thus the product can be rewritten as

$$f(x)=a_n\prod_{j=1}^{l}\rvert x-(a_j+ib_j) \lvert^2$$Now, define a complex-valued polynomial $P(x)$ whose modulus squared gives $f(x)$.

Let
$$P(x)=a_n\prod_{j=1}^{l}(x-(a_j+ib_j))$$and
$$\bar{P(x)}=a_n\prod_{j=1}^{l}(x-(a_j-ib_j))$$When $P(x)$ and $\bar{P(x)}$ are multiplied, the total product includes both $x-(a_j+ib_j)$ and $(x-(a_j-ib_j))$ for each $j$:

$$P(x)\cdot\bar{P(x)}=a_n^2\prod_{j=1}^{l}(x-(a_j+ib_j))(x-(a_j-ib_j)$$And the full product becomes
$$P(x)\cdot\bar{P(x)}=a_n^2\prod_{j=1}^{l}((x-a_j)^2+b_j^2)$$
This matches the original function $f(x)$ up to a constant $a_n^2$ with $2l$ terms.

We can conclude that $$f(x) = \rvert P(x)\lvert^2=P(x)\bar{P(x)}$$ (up to a constant)

The complex polynomial $P(x)$ can be written in terms of its real and imaginary parts as $$P(x)=p(x)+iq(x)$$ where $p(x)$ and $q(x)$ are real polynomials as the real and imaginary parts of $P(x)$. Note that $\bar{P(x)}=p(x)-iq(x)$, hence

$$f(x)=\rvert P(x)\lvert^2=p^2(x)+q^2(x)$$ as required.